Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:21:44Z http://mathoverflow.net/feeds/question/80176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80176/asymptotic-bounds-for-a-confluent-hypergeometric-function-f-11x Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ sigma_z_1980 2011-11-06T00:17:55Z 2011-11-07T11:35:38Z <p>I know that for infinite series and $|z|&lt;1$ there exists a confluent hypergeometric expression </p> <p>$\sum_{k=0}^{\infty} \frac{z^k}{k!k!} = F_{1}[;1;z]$</p> <p>This is not very helpful though, and I 'd like to know if it is possible to get some asymptotic expansion for this function and if there exists some general approach to bounding hypergeometric functions asymptotically. </p> http://mathoverflow.net/questions/80176/asymptotic-bounds-for-a-confluent-hypergeometric-function-f-11x/80185#80185 Answer by Jacques Carette for Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ Jacques Carette 2011-11-06T03:52:11Z 2011-11-06T03:52:11Z <p>This particular function can be expressed in terms of Bessel's I function as $I(0,2\sqrt{z})$, and from there an asymptotic expression (at $\infty$) is easily derived. It starts $$\frac{e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{1}{4}\right)}}{2\sqrt{\pi}} + O\left(e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{3}{4}\right)}\right)$$ where the roots are chosen to have the correct branching behaviour.</p> <p>The easiest way to obtain such results is actually from the ODE satisfied by your function, in this case $zy'' + y' - y$ with $y(0)=1$. It is easy to get an ODE at $\infty$ from there, and from there one gets the asymptotic expansion. The hardest part is getting the singular behaviour 'just right', as well as the branches. Those who have voted to close this likely have never had to compute the asymptotic expansion along a branch cut with the expansion point being an irregular singularity. While it is known how to do this, it is practiced by very very few.</p> http://mathoverflow.net/questions/80176/asymptotic-bounds-for-a-confluent-hypergeometric-function-f-11x/80190#80190 Answer by Brendan McKay for Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ Brendan McKay 2011-11-06T04:49:32Z 2011-11-06T04:49:32Z <p>The OP is unclear about the domain of $z$ required. If only real $z\to\infty$ is of interest, just realise that the terms near $k=\sqrt z$ dominate the rest and their shape is close to a normal density with mean and variance both asymptotically $\sqrt z$. Apply Stirling's formula and that's it.</p> http://mathoverflow.net/questions/80176/asymptotic-bounds-for-a-confluent-hypergeometric-function-f-11x/80200#80200 Answer by S. Sra for Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$ S. Sra 2011-11-06T08:44:35Z 2011-11-06T08:44:35Z <p>Although not directly for $z &lt; 1$, here is a general paper that might help with some of the techniques that one can use:</p> <p><a href="http://dl.acm.org/citation.cfm?id=1655491" rel="nofollow">The confluent hypergeometric functions $M(a; b, z)$ and $U(a; b, z)$ for large $b$ and $z$</a> by J. L. López and P. J. Pagola.</p> <p>Also, if not this paper, please take a look at other papers by López; from what I heard, he is an expert on asymptotics of hypergeometric functions.</p>